Question

Find the value of 2A^3+4A^2, where = \(\begin{bmatrix}5&0&-1\\1&2&-1\\-3&4&1\end{bmatrix}\).

(a) \(\begin{bmatrix}-200&0&-24\\24&-32&-24\\-72&96&-56\end{bmatrix}\)

(b) \(\begin{bmatrix}-200&0&-24\\24&-32&-12\\-72&96&-56\end{bmatrix}\)

(c) \(\begin{bmatrix}-200&0&-24\\12&-32&-24\\-72&96&-56\end{bmatrix}\)

(d) \(\begin{bmatrix}-100&0&-12\\12&-16&-12\\-36&48&-28\end{bmatrix}\)

I had been asked this question in quiz.

The query is from Cayley Hamilton Theorem topic in chapter Eigen Values and Eigen Vectors of Engineering Mathematics

Answer 1

Right choice is (a) \(\begin{bmatrix}-200&0&-24\\24&-32&-24\\-72&96&-56\end{bmatrix}\)

The explanation is: Explanation: For the given Matrix,

A=\(\begin{bmatrix}5&0&-1\\1&2&-1\\-3&4&1\end{bmatrix}\)

The characteristic polynomial is given by-

α^3-(Sum of diagonal elements) α^2+(Sum of minor of diagonal element)α-|A|=0

α^3+2α^2-12α-40=0

The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,

A^3+2A^2-12A+40I=0

A^3+2A^2=12A-40I

A^3+2A^2=\(12\begin{bmatrix}5&0&-1\\1&2&-1\\-3&4&1\end{bmatrix}-40\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

A^3+2A^2=\(\begin{bmatrix}-60-40&0&-12\\12&24-40&-12\\-36&48&12-40\end{bmatrix}\)

A^3+2A^2=\(\begin{bmatrix}-100&0&-12\\12&-16&-12\\-36&48&-28\end{bmatrix}\)

2A^3+4A^2=\(\begin{bmatrix}-200&0&-24\\24&-32&-24\\-72&96&-56\end{bmatrix}\).